Question: Multiply the following complex numbers: $({2-2i}) \cdot ({5})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({2-2i}) \cdot ({5}) = $ $ ({2} \cdot {5}) + ({2} \cdot {0}i) + ({-2}i \cdot {5}) + ({-2}i \cdot {0}i) $ Then simplify the terms: $ (10) + (0i) + (-10i) + (0 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 10 + (0 - 10)i + 0i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 10 + (0 - 10)i - 0 $ The result is simplified: $ (10 - 0) + (-10i) = 10-10i $